3.2711 \(\int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{\sqrt {2+3 x}} \, dx\)

Optimal. Leaf size=160 \[ -\frac {1847 \sqrt {\frac {11}{3}} \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ),\frac {35}{33}\right )}{23625}+\frac {2}{21} (1-2 x)^{3/2} \sqrt {3 x+2} (5 x+3)^{3/2}+\frac {74}{525} \sqrt {1-2 x} \sqrt {3 x+2} (5 x+3)^{3/2}-\frac {1847 \sqrt {1-2 x} \sqrt {3 x+2} \sqrt {5 x+3}}{4725}-\frac {29933 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{23625} \]

[Out]

-29933/70875*EllipticE(1/7*21^(1/2)*(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)-1847/70875*EllipticF(1/7*21^(1/2)*
(1-2*x)^(1/2),1/33*1155^(1/2))*33^(1/2)+2/21*(1-2*x)^(3/2)*(3+5*x)^(3/2)*(2+3*x)^(1/2)+74/525*(3+5*x)^(3/2)*(1
-2*x)^(1/2)*(2+3*x)^(1/2)-1847/4725*(1-2*x)^(1/2)*(2+3*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {101, 154, 158, 113, 119} \[ \frac {2}{21} (1-2 x)^{3/2} \sqrt {3 x+2} (5 x+3)^{3/2}+\frac {74}{525} \sqrt {1-2 x} \sqrt {3 x+2} (5 x+3)^{3/2}-\frac {1847 \sqrt {1-2 x} \sqrt {3 x+2} \sqrt {5 x+3}}{4725}-\frac {1847 \sqrt {\frac {11}{3}} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{23625}-\frac {29933 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{23625} \]

Antiderivative was successfully verified.

[In]

Int[((1 - 2*x)^(3/2)*(3 + 5*x)^(3/2))/Sqrt[2 + 3*x],x]

[Out]

(-1847*Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*Sqrt[3 + 5*x])/4725 + (74*Sqrt[1 - 2*x]*Sqrt[2 + 3*x]*(3 + 5*x)^(3/2))/525
+ (2*(1 - 2*x)^(3/2)*Sqrt[2 + 3*x]*(3 + 5*x)^(3/2))/21 - (29933*Sqrt[11/3]*EllipticE[ArcSin[Sqrt[3/7]*Sqrt[1 -
2*x]], 35/33])/23625 - (1847*Sqrt[11/3]*EllipticF[ArcSin[Sqrt[3/7]*Sqrt[1 - 2*x]], 35/33])/23625

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 113

Int[Sqrt[(e_.) + (f_.)*(x_)]/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-((b*e
- a*f)/d), 2]*EllipticE[ArcSin[Sqrt[a + b*x]/Rt[-((b*c - a*d)/d), 2]], (f*(b*c - a*d))/(d*(b*e - a*f))])/b, x
] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !LtQ[-((b*c - a*d)/d),
0] &&  !(SimplerQ[c + d*x, a + b*x] && GtQ[-(d/(b*c - a*d)), 0] && GtQ[d/(d*e - c*f), 0] &&  !LtQ[(b*c - a*d)
/b, 0])

Rule 119

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*Rt[-(b/d
), 2]*EllipticF[ArcSin[Sqrt[a + b*x]/(Rt[-(b/d), 2]*Sqrt[(b*c - a*d)/b])], (f*(b*c - a*d))/(d*(b*e - a*f))])/(
b*Sqrt[(b*e - a*f)/b]), x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[(b*c - a*d)/b, 0] && GtQ[(b*e - a*f)/b, 0] &
& PosQ[-(b/d)] &&  !(SimplerQ[c + d*x, a + b*x] && GtQ[(d*e - c*f)/d, 0] && GtQ[-(d/b), 0]) &&  !(SimplerQ[c +
d*x, a + b*x] && GtQ[(-(b*e) + a*f)/f, 0] && GtQ[-(f/b), 0]) &&  !(SimplerQ[e + f*x, a + b*x] && GtQ[(-(d*e)
+ c*f)/f, 0] && GtQ[(-(b*e) + a*f)/f, 0] && (PosQ[-(f/d)] || PosQ[-(f/b)]))

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 158

Int[((g_.) + (h_.)*(x_))/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x_Symbol]
:> Dist[h/f, Int[Sqrt[e + f*x]/(Sqrt[a + b*x]*Sqrt[c + d*x]), x], x] + Dist[(f*g - e*h)/f, Int[1/(Sqrt[a + b*
x]*Sqrt[c + d*x]*Sqrt[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && SimplerQ[a + b*x, e + f*x] &&
SimplerQ[c + d*x, e + f*x]

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{3/2} (3+5 x)^{3/2}}{\sqrt {2+3 x}} \, dx &=\frac {2}{21} (1-2 x)^{3/2} \sqrt {2+3 x} (3+5 x)^{3/2}-\frac {2}{21} \int \frac {\left (-30-\frac {111 x}{2}\right ) \sqrt {1-2 x} \sqrt {3+5 x}}{\sqrt {2+3 x}} \, dx\\ &=\frac {74}{525} \sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}+\frac {2}{21} (1-2 x)^{3/2} \sqrt {2+3 x} (3+5 x)^{3/2}-\frac {4 \int \frac {\left (-\frac {1503}{4}-\frac {5541 x}{4}\right ) \sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx}{1575}\\ &=-\frac {1847 \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}}{4725}+\frac {74}{525} \sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}+\frac {2}{21} (1-2 x)^{3/2} \sqrt {2+3 x} (3+5 x)^{3/2}+\frac {4 \int \frac {\frac {119949}{8}+\frac {89799 x}{4}}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{14175}\\ &=-\frac {1847 \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}}{4725}+\frac {74}{525} \sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}+\frac {2}{21} (1-2 x)^{3/2} \sqrt {2+3 x} (3+5 x)^{3/2}+\frac {20317 \int \frac {1}{\sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}} \, dx}{47250}+\frac {29933 \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x} \sqrt {2+3 x}} \, dx}{23625}\\ &=-\frac {1847 \sqrt {1-2 x} \sqrt {2+3 x} \sqrt {3+5 x}}{4725}+\frac {74}{525} \sqrt {1-2 x} \sqrt {2+3 x} (3+5 x)^{3/2}+\frac {2}{21} (1-2 x)^{3/2} \sqrt {2+3 x} (3+5 x)^{3/2}-\frac {29933 \sqrt {\frac {11}{3}} E\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{23625}-\frac {1847 \sqrt {\frac {11}{3}} F\left (\sin ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )|\frac {35}{33}\right )}{23625}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 97, normalized size = 0.61 \[ \frac {1085 \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right ),-\frac {33}{2}\right )+15 \sqrt {2-4 x} \sqrt {3 x+2} \sqrt {5 x+3} \left (-4500 x^2+2880 x+1501\right )+59866 E\left (\sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )|-\frac {33}{2}\right )}{70875 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - 2*x)^(3/2)*(3 + 5*x)^(3/2))/Sqrt[2 + 3*x],x]

[Out]

(15*Sqrt[2 - 4*x]*Sqrt[2 + 3*x]*Sqrt[3 + 5*x]*(1501 + 2880*x - 4500*x^2) + 59866*EllipticE[ArcSin[Sqrt[2/11]*S
qrt[3 + 5*x]], -33/2] + 1085*EllipticF[ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]], -33/2])/(70875*Sqrt[2])

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fricas [F]  time = 1.27, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (10 \, x^{2} + x - 3\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{\sqrt {3 \, x + 2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)^(3/2)/(2+3*x)^(1/2),x, algorithm="fricas")

[Out]

integral(-(10*x^2 + x - 3)*sqrt(5*x + 3)*sqrt(-2*x + 1)/sqrt(3*x + 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (-2 \, x + 1\right )}^{\frac {3}{2}}}{\sqrt {3 \, x + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)^(3/2)/(2+3*x)^(1/2),x, algorithm="giac")

[Out]

integrate((5*x + 3)^(3/2)*(-2*x + 1)^(3/2)/sqrt(3*x + 2), x)

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maple [C]  time = 0.01, size = 150, normalized size = 0.94 \[ -\frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \left (4050000 x^{5}+513000 x^{4}-4283100 x^{3}-1240890 x^{2}+833610 x +59866 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticE \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+1085 \sqrt {2}\, \sqrt {5 x +3}\, \sqrt {3 x +2}\, \sqrt {-2 x +1}\, \EllipticF \left (\frac {\sqrt {110 x +66}}{11}, \frac {i \sqrt {66}}{2}\right )+270180\right )}{141750 \left (30 x^{3}+23 x^{2}-7 x -6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x+1)^(3/2)*(5*x+3)^(3/2)/(3*x+2)^(1/2),x)

[Out]

-1/141750*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(3*x+2)^(1/2)*(1085*2^(1/2)*(5*x+3)^(1/2)*(3*x+2)^(1/2)*(-2*x+1)^(1/2)*
EllipticF(1/11*(110*x+66)^(1/2),1/2*I*66^(1/2))+59866*2^(1/2)*(5*x+3)^(1/2)*(3*x+2)^(1/2)*(-2*x+1)^(1/2)*Ellip
ticE(1/11*(110*x+66)^(1/2),1/2*I*66^(1/2))+4050000*x^5+513000*x^4-4283100*x^3-1240890*x^2+833610*x+270180)/(30
*x^3+23*x^2-7*x-6)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (-2 \, x + 1\right )}^{\frac {3}{2}}}{\sqrt {3 \, x + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)^(3/2)*(3+5*x)^(3/2)/(2+3*x)^(1/2),x, algorithm="maxima")

[Out]

integrate((5*x + 3)^(3/2)*(-2*x + 1)^(3/2)/sqrt(3*x + 2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-2\,x\right )}^{3/2}\,{\left (5\,x+3\right )}^{3/2}}{\sqrt {3\,x+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - 2*x)^(3/2)*(5*x + 3)^(3/2))/(3*x + 2)^(1/2),x)

[Out]

int(((1 - 2*x)^(3/2)*(5*x + 3)^(3/2))/(3*x + 2)^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-2*x)**(3/2)*(3+5*x)**(3/2)/(2+3*x)**(1/2),x)

[Out]

Timed out

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